Exponential Growth and Decay, Rates of Change, and the Limits of a Model

A deer population in a nature reserve is modelled by $N = 340(1.06)^t$, where $N$ is the number of deer and $t$ is the number of years since the reserve was established. What was the population when the reserve was established?

• $340$

  Correct. "When the reserve was established" is $t=0$, and $(1.06)^0=1$, so $N(0)=340$ — read directly with no method needed, exactly what spec 4.4's own use of the word "initial" means.

• B$360$ (to the nearest whole deer) — the population after one year

  This is $N(1) = 340\times1.06 \approx 360$, one year AFTER establishment, not at $t=0$. "Initial" means the very first instant the model describes, not the first period of data that follows it.

• C$1.06$

  This is the growth multiplier — the base of the exponential term, controlling how the population changes from one year to the next. It is not a population count at all, and reading it as one confuses the coefficient in front of the exponential with the base inside it.

• D$6\%$

  This is the annual percentage growth rate implied by the base $1.06$ ($1.06 = 1+0.06$) — a genuinely meaningful number, but a rate of change, not a population count, and not what "initial" refers to at all.

Traps tested: Initial value confused with first period value · Growth rate confused with initial value

What is $\frac{d}{dx}\left[6(2.5)^x\right]$?

• $6(2.5)^x\ln(2.5)$

  Correct. Applying $\frac{d}{dx}[a^x]=a^x\ln a$ directly, with the constant $6$ carried through by the constant-multiple rule.

• B$6(2.5)^x$

  The $\ln(2.5)$ multiplier has been left out entirely — as though $a^x$ differentiated to itself, unchanged. That would only be true for $\ln a=1$, i.e. $a=e$; for every other base, the multiplier is genuinely there.

• C$12x(2.5)^{x-1}$

  This applies the power rule ($nx^{n-1}$) as though $x$ were the base and $2.5$ the fixed exponent — the opposite shape to $6(2.5)^x$, which has a constant base and $x$ in the exponent. Different shapes need different rules: this one needs $a^x\to a^x\ln a$, not the power rule.

• D$6(2.5)^x\ln(6)$

  The multiplier is $\ln$ of the wrong number — it should be $\ln$ of the BASE, $2.5$, not $\ln$ of the coefficient sitting in front, $6$. The mechanism this lesson derives shows exactly where the $\ln$ comes from: rewriting the base $a^x$ as $e^{x\ln a}$, which only ever involves $\ln$ of the base.

Traps tested: Ln a multiplier omitted · Power rule applied to exponential function · Ln of coefficient used instead of ln of base

A population of a native fish species in a lake, monitored after an invasive species was introduced, is modelled by $P=2000(0.9)^t$, where $P$ is the number of fish and $t$ is the number of years since monitoring began. Find the exact rate at which the population is decreasing when $t=10$, to 3 significant figures.

• $73.5$ fish per year (decreasing)

  Correct. $\frac{dP}{dt}=2000(0.9)^t\ln(0.9)$; at $t=10$, $\frac{dP}{dt}\approx-73.5$. The negative sign means DEcreasing, so the rate of decrease is $73.5$ fish per year (3 s.f.) — the unit and the direction word both present.

• B$697$ fish

  This is $P(10)$, the population VALUE at $t=10$ — substituted directly with no differentiation at all. This is the exact real, documented error: "substituted t = ... directly into N... [earned] no marks" (Jun 2023 Q7(b)) — a completely different quantity from the rate being asked for, not merely an inaccurate version of it.

• C$69.7$ fish per year (decreasing)

  This is the AVERAGE rate of change over the year from $t=10$ to $t=11$, $P(11)-P(10)$ — a genuinely close-looking number to the correct $73.5$, but still a finite difference, not the instantaneous rate differentiation finds at the single point $t=10$. Closeness doesn't earn credit here; it's a documented, separately named error pattern from the same source.

• D$73.5$ fish per year (increasing)

  The magnitude is exactly right and the direction is backwards. $\frac{dP}{dt}$ comes out NEGATIVE here because $\ln(0.9)<0$ (the base is less than 1, a decaying model) — reporting the same correct number with the wrong direction word is exactly the kind of error the units/context-word trap in this lesson documents, just with the direction flipped rather than omitted.

Traps tested: Value substituted instead of rate found · Average rate substituted for instantaneous rate · Sign of rate not interpreted

A tank is being drained, and the volume of water remaining is modelled by $W=500(0.75)^t$, where $W$ is in litres and $t$ is in minutes. A student correctly finds $\frac{dW}{dt}\big|_{t=2} \approx -80.9$, and writes their final answer as simply "$80.9$". Which of the following best describes this answer?

• The number is right, but the answer is incomplete — it needs the unit (litres per minute) and a statement that the tank is draining

  Correct. $80.9$ is genuinely the correct magnitude (3 s.f.) — the arithmetic that produced it is not in question. What's missing is exactly the documented trap: no unit attached, and no statement of direction. A mark scheme treats this as an incomplete answer, not as a correct one written briefly.

• BThe number itself is wrong — it should be $281$

  $281$ is $W(2)$, the volume remaining at $t=2$ — a different quantity from the rate of drainage, and not what's being questioned here. $80.9$ genuinely is the correct rate; the problem is what's missing around it, not the number itself.

• CThe answer is fully correct as written — units are only needed if the question explicitly asks for them

  Units and context words for a modelling answer are not optional extras contingent on the question asking twice — This course's own research documents this exact omission costing marks across at least four separate series, with correct numbers attached every time. A bare number answering a real-world rate question is genuinely incomplete.

• DThe answer is fully correct, since the sign was correctly interpreted as positive

  There's no statement of direction anywhere in "$80.9$" for anything to have been correctly interpreted — the number alone doesn't say whether the tank is filling or draining. That missing statement is precisely the gap, not a sign that's been silently handled correctly.

Traps tested: Correct rate and wrong value conflated · Unit treated as optional presentation

A perfume's scent intensity is modelled by $S=5+45(0.8)^t$ (on a 0–100 scale), $t$ hours after application, approaching a faint background level of $5$ as $t\to\infty$. Which of the following correctly explains why $S=3$ has no solution for $t\geq0$?

• Solving would require $(0.8)^t = -\frac{2}{45}$, and $(0.8)^t$ is strictly positive for every real $t$, so it can never equal a negative number

  Correct. Rearranging $S=3$ gives $45(0.8)^t = 3-5=-2$, so $(0.8)^t=-\frac{2}{45}$ — impossible, because an exponential term with a positive base is never zero or negative. The model's asymptote sits at $S=5$, approached from above, so $S$ can never reach $5$, let alone fall below it to $3$.

• BBecause $(0.8)^t$ is always less than $1$ for $t>0$

  True on its own, but not the actual reason the equation fails — $(0.8)^t$ being less than 1 is perfectly consistent with plenty of solvable equations. The real obstruction is that solving $S=3$ would need $(0.8)^t$ to be NEGATIVE, which no exponential with a positive base can ever be, whatever size it is.

• CBecause the model is only valid for whole-number values of $t$

  Nothing in the model or the question restricts $t$ to whole numbers — this introduces a restriction that was never stated, rather than identifying the genuine one (the positivity of the exponential term) that actually rules out $S=3$.

• DBecause $45$ is larger than $5$

  This compares two of the model's constants with no connection to why the specific equation $S=3$ fails — it isn't a step in any valid reasoning about the domain of this model.

Traps tested: True but irrelevant fact substituted for the actual mechanism · Unstated restriction invented · Unrelated numerical comparison substituted for reasoning

A model $V=24000(0.88)^t$ predicts the value of a piece of equipment continuing to fall towards £0 for large $t$. What is the strongest reason to prefer a second, improved model with a floor at the equipment's £2000 scrap value, for predictions beyond the range the original data covered?

• Real equipment retains a recoverable scrap or resale value and does not become genuinely worthless, so a model whose long-run prediction is £0 is unrealistic

  Correct — this is exactly spec 4.4's own guidance on "a second, improved model": the original model's own long-run mathematical behaviour (V→0) is checked against what's actually plausible for the real quantity being modelled, and found wanting.

• BBecause $0.88$ is close to $1$, so the model decays only slowly

  This is a true observation about the SPEED of decay, unrelated to whether the model's long-run DESTINATION (£0) is realistic. A model can decay slowly or quickly and still be heading somewhere unrealistic — the two are separate questions.

• CBecause the original model is mathematically invalid for $t>12$

  There's nothing mathematically invalid about $24000(0.88)^t$ for any $t\geq0$ — it's a perfectly well-defined, positive, decreasing function everywhere. The issue is realism, not validity: the model computes a value fine, it's just a value that stops matching reality.

• DBecause the original model gives a negative value for some $t$ in $0\leq t\leq 12$

  It doesn't — $24000(0.88)^t$ is strictly positive for every real $t$, the same fact established earlier in this lesson for any exponential term with a positive base. This distractor tests whether the reasoning is genuinely understood or just pattern-matched to "something about large t going wrong."

Traps tested: Decay speed confused with long run realism · Unrealistic prediction confused with mathematical invalidity · False premise invented to justify correct conclusion

A drug's concentration remaining in a patient's bloodstream is modelled by $D=250(1.15)^{-t}$, where $D$ is in mg and $t$ is the number of hours since the drug was administered — the same shape as a real WMA13 question (Jan 2022 Q8, an antibiotic-decay model with a base greater than $1$ and a NEGATIVE exponent, rather than a base less than $1$). What is $\frac{dD}{dt}$?

• $-250\ln(1.15)(1.15)^{-t}$

  Correct. Rewrite the exponent as $-t=k t$ with $k=-1$: the chain rule (already established as a prerequisite) means $\frac{d}{dt}\left[(1.15)^{-t}\right]=(1.15)^{-t}\ln(1.15)\times(-1)$, an EXTRA factor of $-1$ beyond the $a^t\to a^t\ln a$ result alone. The constant $250$ carries through unchanged. The negative sign genuinely belongs here — this model is decaying (concentration falling as the drug clears), so its derivative should be negative — and dropping it is the single most consistently documented differentiation slip for exactly this shape of model (see the trap taxonomy above).

• B$250\ln(1.15)(1.15)^{-t}$

  This is the exact real, documented error for this shape of model: the chain-rule factor of $-1$ from differentiating the exponent $-t$ has been dropped, leaving a POSITIVE derivative for a model that is genuinely decreasing. Verified verbatim from the real Jan 2024 Q5(c) mark scheme, for the equivalent slip: "If they lose the minus sign in ...e−Bt they obtain ±0.0518… and this scores M0" — in a real mark scheme this sign is folded directly into the method mark, so losing it can cost the entire part.

• C$-250t(1.15)^{-t-1}$

  This applies the power rule ($nx^{n-1}$) to $-t$ as though it were the base with a fixed exponent — the opposite shape to $250(1.15)^{-t}$, which has a constant base ($1.15$) and $t$ in the exponent. Verified verbatim, Jan 2022 Q8(c) examiner report, for exactly this confusion: "Common errors were to miss off the negative sign and treated the t as a constant power and subtracting 1 from it (this incorrect method achieved no marks for this part)."

• D$-250(1.15)^{-t}$

  The $\ln(1.15)$ multiplier has been left out entirely, as though $a^{-t}$ differentiated to $-a^{-t}$ with no rate constant attached at all. The sign is right here, but the magnitude is missing its own scaling factor — every base except $a=e$ needs its own $\ln a$ multiplier, exactly as the mechanism derived earlier in this lesson shows.

Traps tested: Negative exponent sign dropped during differentiation · Power rule applied to exponential function · Ln a multiplier omitted